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\author{五六七 }
\title{最速下降线的方程 }

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\begin{document}

\maketitle

\begin{abstract}
建立最速下降线符合的微分方程，求解函数表达式。并用模拟数据进行比较。
\end{abstract}

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\section{问题描述}
在高低不同的两个点之间铺设一条坡道，使得小球在重力作用下沿着坡道从高点静止下滑，到达低点所用的时间最短。

\begin{figure}[ht!]\centering
\includegraphics [height=4cm, width=7cm]{brachistochrone.png}
\caption{从高点到低点的一条坡道 }
\end{figure}




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\section{内容提要}
\begin{itemize}
\item 泰勒斯定理，伽利略运动定律，弦定律。
\item 最速下降线、等时线和旋轮线的概念。
\item 费马最短时间原理，折射公式，伯努利求出最速下降线的方法。
\end{itemize}

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\section{相关问题}
\begin{enumerate}
\item 证明弦定律。
\item 什么是最速下降线、等时线和旋轮线？
\item 证明旋轮线下方的区域面积正好等于旋轮面积的三倍。
\item 用费马原理证明折射原理。
\item 约翰伯努利是怎样计算最速下降线的？
\item 雅克布伯努利是怎样计算最速下降线的？
\item 编写程序模拟小珠沿着给定曲线下降的过程，并计算所用的时间。

\end{enumerate}

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\section{建立微分方程模型}

\begin{enumerate}\setcounter{enumi}{0}\itemsep0.5em

\item   设所求曲线的方程为 $x=x(y)$, 纵坐标 $y$ 正向向下，横坐标 $x$ 正向向右。
\item  我们要得出函数 $x(y)$ 所满足的微分方程。

\begin{figure}
 \centering
 \includegraphics[height=5cm, width=9cm]{brachistochrone-2.png}
 %\caption{ }
 \end{figure}

\item  在下降高度为 $y$ 的任何地方，势能 $mgy$ 全都转化成动能 $\frac{1}{2}mv^2$, 因此有
\begin{eqnarray*}
\frac{1}{2}mv^2 &=& mgy, \\ 
v &=& \sqrt{2gy}. 
\end{eqnarray*}

\item  根据折射定律，在两种介质中的光线，入射角与出射角的正弦比，等于光线在两种介质中的速度比。由此可得
\begin{eqnarray*}
\frac{\sqrt{y}}{\sqrt{y+dy}} =  \frac{ \frac{dx_1}{\sqrt{dy^2+dx_1^2}} } {\frac{dx_2}{\sqrt{dy^2+dx_2^2}}}. 
\end{eqnarray*}

\item  两边平方，整理可得
\begin{eqnarray*}
\frac{y+dy}{y}  = \frac{dx_2^2(dy^2+dx_1^2)}{dx_1^2(dy^2+dx_2^2)}. 
\end{eqnarray*}

\item  两边减去1，可得
\begin{eqnarray*}
\frac{dy}{y}  = \frac{dy^2(dx_2^2-dx_1^2)}{dx_1^2(dy^2+dx_2^2)} = \frac{dy^2(dx_2+dx_1)(dx_2-dx_1)}{dx_1^2(dy^2+dx_2^2)} . 
\end{eqnarray*}

\item  移项整理可得  
\begin{eqnarray*}
\frac{dydx_1^2(dy^2+dx_2^2)}{ydy^4(dx_2+dx_1)}  = \frac{dx_2-dx_1}{dy^2} . 
\end{eqnarray*}

\item  使用 $dx_1\approx dx_2\approx dx$, 以及 $\frac{dx_2-dx_1}{dy^2} = \frac{d^2x}{dy^2}$, 化简可得 
\begin{eqnarray*}
\frac{1}{2y}\frac{dx}{dy}\left[1+ \left(\frac{dx}{dy} \right)^2 \right] = \frac{}{}\cdot \frac{d^2x}{dy^2} 
\end{eqnarray*}

\item  这是一个关于未知函数 $x=x(y)$ 的二阶微分方程 
\begin{eqnarray*}
x'' = \frac{x'(1+x'^2)}{2y}. 
\end{eqnarray*}


\end{enumerate}

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\section{求解微分方程}

\begin{enumerate}\setcounter{enumi}{0}\itemsep0.5em

\item  设 $z=\frac{dx}{dy}$, 则得到一阶微分方程
\begin{eqnarray*}
2yz' = z(1+z^2). 
\end{eqnarray*}

\item  使用分离变量法，可得
\begin{eqnarray*}
\frac{dz}{z(1+z^2)} = \frac{dy}{2y}.  
\end{eqnarray*}

\item  两边积分，可得
\begin{eqnarray*}
\ln z -\frac{1}{2}\ln (1+z^2) = \frac{1}{2}\ln y + C, 
\end{eqnarray*}
其中 $C$ 是任意常数。

\item  整理可得如下方程，其中 $k^2$ 是任意常数，
\begin{eqnarray*}
z = \sqrt { \frac{y}{k^2-y} }. 
\end{eqnarray*}

\item  因此所求的最速下降线满足一阶微分方程
\begin{eqnarray*}
\frac{dx}{dy} = \sqrt { \frac{y}{k^2-y} }. 
\end{eqnarray*}

\item  分离变量，两边积分，设起点为坐标原点，可得
\begin{eqnarray*}
x = \int_0^y \sqrt { \frac{t}{k^2-t} }dt  
\end{eqnarray*}

\item  根据 \cite{wolfram}, 验证可知参数方程的解函数
\begin{eqnarray*}
\left\{\begin{array}{rcl}
x &=& \frac{1}{2}k^2(\theta - \sin\theta), \\ 
y &=& \frac{1}{2}k^2(1-\cos\theta), 
\end{array}\right. 
\end{eqnarray*}
满足微分方程
\begin{eqnarray*}
\frac{dx}{dy} = \sqrt { \frac{y}{k^2-y} }. 
\end{eqnarray*}



\end{enumerate}


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\section{编程计算}




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\section{回答问题}

最速下降线是一条旋轮线。 

\begin{figure}
 \centering
 \includegraphics[height=5cm, width=9cm]{brachistochrone-3.png}
 %\caption{ }
 \end{figure}


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\section{实验测量}



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\newpage

\section{习题}

用费马原理（光线总是最短时间的路径前进）证明光线的折射原理：光线在两种介质中的路径发生偏折，入射角和出射角的正弦的比例，等于光线在两种介质中的速度的比例，即 $$\frac{\sin\alpha_1}{\sin\alpha_2} = \frac{v_1}{v_2}. $$

\begin{figure}[ht!]
 \centering
 \includegraphics[height=5cm, width=9cm]{fermat-refraction-snell-law.png}
 %\caption{ }
 \end{figure}
 
 
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%\section{参考文献 }
\begin{thebibliography}{99}
\bibitem{dingtongren} 丁同仁、李承治，\emph{常微分方程教程}，高等教育出版社，2022年3月第三版。
\bibitem{sishoukui-2} 司守奎,孙玺菁. \emph{Python数学建模算法与应用}, 国防工业出版社. 2022年1月第1版. 
\bibitem{jiangqiyuan} 姜启源, 谢金星, 叶俊. \emph{数学模型}, 高等教育出版社. 2018年5月第5版. （第13.1节）
\bibitem{babb-currie} Jeff Babb and James Currie. \emph{The Brachistochrone Problem: Mathematics for a Broad Audience via a Large Context Problem}. The Montana Mathematics Enthusiast, ISSN 1551-3440, Vol. 5, nos.2\&3, pp.169-184.
\bibitem{wolfram} \url{https://mathworld.wolfram.com/BrachistochroneProblem.html}. 

\end{thebibliography}

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\end{document}

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